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104.A bar fixed at its upper end is loaded by a longitudinal force P. There is a clearance between

the lower end of the bar and the lower rigid support. When P EA/l, the lower clearance is taken

up. Find the reaction N of the lower support.

If P1 is the load required to elongate upper portion by , then = P1l/EA ...................(1)

If P > P1, then an additional load (P-P1) acts at section B,(causing compression of bottom portion and

further extension of top) once the bottom of lower portion of the rod touches the lower rigid support

( P1 load is used up for extending top portion of the rod by )

Now both top and bottom portion are subjected to compressive load N, where N is the

reaction from lower rigid support.

(Note that (P1 + (P-P1)/2) + ((P-P1)/2 ) = P)

For equilibrium of the lower portion: N = P-P1/2 ......................................(2)

Also N or (P-P1)/2= EA/l ....................................................................................(3)

Fig 82.

Fig 83.

(a)(b)

(c)(d)

l

l

P

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From(2), 2N = P-P1 P1 = P-2N

from(1) = l/EA(P-2N)

EA/l = P-2N 2N = P- EA/l

or N = P/2 (/l) .(EA/2) ......................................................................................(4)

105. In the above problem, we have proved that

N = P/2 (/l) (EA/2)

The upper part is extended by the force P-N

ie.,P-N = P - (P/2 - (/l) (EA/2)) = P/2 + (/l) (EA/2) .........................................(5)

The displacement (+) of the point of application of the force P

(P-N)l/EA = (l/EA)[P/2 -/l (EA/2)]

ie., (+) = Pl/2EA+/2 ........................................................................................(6)

We determine the elastic energy stored in the bar. On one hand, this energy may be determined as the

sum of the energies contained in the upper and lower portions of the bar.

ie.,U = (P-N)2l/(2EA) + N2l/2EA

or U = [P/2 +(/l) (EA/2)]2l/2EA + N2l/2EA, by(5)

= [P/2 +/l (EA/2)]2l/2EA + (l/2EA) ([P/2 (/l) (EA/2)]2) by(4)

= l/2EA[2{ P2/4 +2/l2 (E2A2/4)}]

U = P2l/4EA + 2EA/4l ....................................................................................(7)

On the other hand, this energy is equal to the work done by the force P during the displacement (+

) ie.,

U = P (+) P/2[Pl/2EA + /2], by (6)

U = P2l/4EA + P /4 .............................................................................................(8)

But expressions (7) and (8) are different. What is the matter? Which of these expressions are correct

and which is not?

Expression (7) is correct and expression (8) is erroneous. In the present case W.D. by the force P

U P (+) since the displacement (+) is not proportional to the force P over the entire portion.

This can easily be seen from the graph of figure 84 which shows the relation (+) and P for

the system under consideration.

P

EA/L

LOAD

DEFLECTION

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The W.D. by the force P is determined by the shaded area.

ie., U = (EA/l). +[ (P- EA/l) + (EA/l)x]

= (EA2/2l) + .[Pl/2EA+/2-][P- EA/l] + (EA/l){ Pl/2EA-/2}

= (EA2/2l) + .[Pl/2EA-/2][P- EA/l] + (EA/l){ Pl/2EA-/2}

= (EA2/2l) +[P2l/2EA-P/2 - P/2+EA2/2l] + P/2-EA2/2l

= (EA2/2l) +[P2l/2EA-P+EA2/2l] + P/2 - EA2/2l

= P2l/4EA + EA2/4l

This expression is identical with that obtained previously.

106. A homogeneous straight beam of length l and weight P rests on a rigid plane. If a force P1(

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After applying force P1, the left hand part of the

beam rises a little over a certain length a. The

right-hand part continues to lie on the ground

and remains straight. Consequently, at all

sections of the right-hand portion the bending

moment is zero. In particular, the moment is zeroat the section x = a as well.

This condition determines the length of the

segment a,

ie., P1 a = q a2/2

= Pa2/2l

a=(2P1/P)l

from the condition that the sum of projections of

all forces on a vertical axis is zero it follows that

the rigid plane produces a reaction P1 at the pointx=a.

With this system of forces, the left-hand

overhanging part of the beam may be regarded

as a simply supported beam of length a

subjected to a u.d.l of intensity P/l. The

maximum B.M occurs at the middle of the

overhanging portion and is equal to :

q a2/8 = P12l/2P

Note: In solving this problem, we considered

only the flexural rigidity of the beam and

assumed no shearing strains at the cross sections.

So this is an approximate solution. Actually

force P1 is not concentrated force but is

distributed over a small area.

107. A composite bar of steel and brass is heated through some temperature. Which of the two bars

will be subjected to tensile force and which to compressive force and why?

Fig 85.

(a)

(b)

(c)

Fig 86.

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Since the bar is composite, the members are rigidly fixed, so the composite bar, as a whole,

expands by the same amount. But brass expands more than steel ( the coefficient of expansion of

brass > coefficient of expansion of steel) and therefore free expansion of brass will be more than that

of steel. But, since both members are not free to expand, therefore the expansion of composite bar, asa whole, will be less than that of brass; but more than that of steel.

It is therefore obvious, that the brass will be subjected to compressive force, whereas the steel

will be subjected to tensile force.

108. What are the components (figure 88) of force P in the directions OA and OB.

Compete the parallelogram OCPD as shown above.

OC is component of force P along OA

& OD is component of force P along OB.

Note: Component of force along OA Pcos

And Component of force along OA Psin

109. What do you mean by the terms flexibility and stiffness with regard to a bar?

The flexibility of the bar is defined as the deflection due to a unit value of load; for a bar it is l/EA.

( = Pl/EA)

The stiffness of the bar is defined as the force required to produce a unit deflection; thus, the

stiffness is equal to EA/L [ P = (EA/L). ] which is the reciprocal of flexibility.

110. Find the reactions Ra, Rb for the statically indeterminate bar shown in figure 89 by forcemethod(or flexibility method).

Fig 87.

Fig 88.(a) (b)

sB

S B S B S B

B

P

C

O

D

A

O

A

B

P

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We select one of the reactions, say Ra, as the unknown quantity. Rb can be found from equilibrium

equations:

Ra + Rb = P ....................(1)

Unknown Ra is called statically redundant, if it is removed,(see figure 89b) the structure

that remains is statically determinate and stable.

The structure that remains after releasing the redundant is called released structure or primary

structure (figure 89b). For this structure, displacement of point A: p = Pb/EA.....................(2)

is downward. In figure 89(c) we consider the effect of redundant Ra on the displacement of point A;

Ra is visualized as load acting on the released structure. IfR is the upward displacement of point A

due to load Ra then

R= Ra L/EA ................(3)

The final displacement of point A due to both P and Ra acting simultaneously is given by

= P - R But of point A is zero.

P = R Ra L/EA = Pb/EA (compatability of displacements)

Ra = Pb/L

from(1), Rb = Pa/L, Thus both reactions have been found.

111. Solve the previous problem, by displacement method of analysis.

Fig 89.

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In this method we take displacement c of point C, the junction of two parts of the bar, as the

unknown quantity.

The forces Ra and Rb in the top and bottom parts of the bar can be expressed in terms of c, as

follows:

Ra = (EA/a)c ; Rb = (EA/b)c ...................(1)

Here we have assumed that c is positive downward, so that tension is produced in the upper part &

compression is produced in the lower part of the bar.

Next step is to isolate point c in the bar as a free body (figure 90b) and consider its equilibrium.

For this free body,

Ra + Rb = P ................................................(2)

From (1), (EA/a)c + (EA/b)c = P

c = (Pab/EAL) ....................................(3)

Knowing c, we can find Ra and Rb from equations (1):

Ra = Pb/L; Rb = Pa/L.

Fig 90.